Prime factorization of n+1
Check out the literature on Fermat numbers, $2^{2^n}+1$. If factoring $m$ helped you factor $m+1$, these numbers would be a cinch, but they're not.
The general philosophy is that multiplication and addition do not "see" each other. So the fact that one knows the multiplicative structure of n does not say anything about the multiplicative structure of n+1. There are several demonstrations of this philosophy.
One of them concerns twin primes: a well-known heuristic, first exploited by Cramer, says that a random number $n$ is prime with probability $1/\log n$ (this is supported by the Prime Number Theorem). However, if we assume that $n$ is a prime number, then it is believed that the probability of $n+2$ being a prime number is still $1/\log n$, provided that there no local obstructions to this (for example, if $n\equiv1\pmod 3$, then this is trivially false). In other words, $n+2$ does not "know" whether $n$ is prime or not. Indeed, a quantitative form of the twin prime conjecture states that
$$|\{n\le x:n~{\rm and}~n+2~{\rm are~prime~numbers}\}|\sim\frac{cx}{\log^2x},$$
where $c$ is some constant which arises due to the local obstructions mentioned above.
A second demonstration of this philosophy is the Erdos-Szemeredi conjecture which, in its simplest form, states that if $A$ is a set of integers and we set
$$A+A=\{a+b:a,b\in A\}\quad{\rm and}\quad A\cdot A=\{a\cdot b:a,b\in A\},$$
then
$$\max\{|A+A|,|A\cdot A|\}\ge c_\epsilon|A|^{2-\epsilon}$$
for every $\epsilon>0$, where $c_\epsilon$ is some constant that depends only on $\epsilon$. Roughly, this conjecture says that $A$ cannot have both additive and multiplicative structure, which would reduce the cardinality of $A+A$ and $A\cdot A$.
To elaborate on azorne's answer. We can do it in a way reminding of how we can take $n$'th powers modulo a number in about $\log n$ time.
Assume that there is a fast way to do what you want, and that we want to factor $n$. Then either $n$ or $n-1$ is divisible by 2. If $n-1$ is divisible by 2 then this reduces down to factor $(n-1)/2$ + one operation of knowing the factorization of $n-1$ to obtain a factorization of $n$. If $n$ is divisible by two we can just divide by two to reduce the factorization to the factorization of $n/2$.
Thus we see that to factor $n$ takes at most the time to factor $[n/2]$ + One operation of knowing the factor of $n-1$ to factor $n$.
If we do this in $\log_2 n$ steps we will come down to trivial numbers to factor, and thus we see that the time it takes to factor a number $n$ will be at most $\log_2 n \times $ "the maximum time it takes to go from knowing the factorization of $m$ to factor $m+1$ for $m \leq n$". This can certainly not be fast (e.g. polynomial time in $\log n$) since it would give a polynomial time algorithm (in $\log n$) to factor an arbitrary number $n$. No such algorithm is known of course. The number field sieve is expected to be the fastest known algorithm discovered yet, although the estimates for the time complexity it takes is just estimated heuristicly and is not proven rigorously.